CSC 413/513: Intro to Algorithms

CSC 413/513: Intro to Algorithms Hash Tables

Hash Tables • Hash table: • Given a table T and a record x, with key (= symbol) and satellite data, we need to support: • Insert (T, x) • Delete (T, x) • Search(T, x) • We want these to be fast, but don’t care about sorting the records • In this discussion we consider all keys to be (possibly large) natural numbers

Direct Addressing • Suppose: • The range of keys is 0..m-1 • Keys are distinct • The idea: • Set up an array T[0..m-1] in which • T[i] = x if x T and key[x] = i • T[i] = NULL otherwise • This is called a direct-address table • Operations take O(1) time!

The Problem With Direct Addressing • Direct addressing works well when the range m of keys is relatively small • But what if the keys are 32-bit integers? • Problem 1: direct-address table will have 232 entries, more than 4 billion • Problem 2: even if memory is not an issue, the time to initialize the elements to NULL may be • Solution: map keys to smaller range 0..m-1 • This mapping is called a hash function

Hash Functions • Next problem: collision T 0 U(universe of keys) h(k1) k1 h(k4) k4 K(actualkeys) k5 h(k2) = h(k5) k2 h(k3) k3 m - 1

Resolving Collisions • How can we solve the problem of collisions? • Solution 1: chaining • Solution 2: open addressing

Open Addressing • Basic idea (details in Section 11.4): • To insert: if slot is full, try another slot, …, until an open slot is found (probing) • To search, follow same sequence of probes as would be used when inserting the element • If reach element with correct key, return it • If reach a NULL pointer, element is not in table • Good for fixed sets (adding but no deletion) • Example: spell checking • How to accommodate deletion? What is the impact? • Table needn’t be much bigger than n

Chaining • Chaining puts elements that hash to the same slot in a linked list: n can be larger than m T —— U(universe of keys) k1 k4 —— —— k1 —— k4 K(actualkeys) k5 —— k7 k5 k2 k7 —— —— k3 k2 k3 —— k8 k6 k8 k6 —— ——

Chaining • How do we insert an element? T —— U(universe of keys) k1 k4 —— —— k1 —— k4 K(actualkeys) k5 —— k7 k5 k2 k7 —— —— k3 k2 k3 —— k8 k6 k8 k6 —— ——

Chaining • How do we delete an element? • Do we need a doubly-linked list for efficient delete? T —— U(universe of keys) k1 k4 —— —— k1 —— k4 K(actualkeys) k5 —— k7 k5 k2 k7 —— —— k3 k2 k3 —— k8 k6 k8 k6 —— ——

Chaining • How do we search for an element with a given key? T —— U(universe of keys) k1 k4 —— —— k1 —— k4 K(actualkeys) k5 —— k7 k5 k2 k7 —— —— k3 k2 k3 —— k8 k6 k8 k6 —— ——

Analysis of Chaining • Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot • Given n keys and m slots in the table: the load factor = n/m = average # keys per slot • What will be the average cost of an unsuccessful search for a key?

Analysis of Chaining • Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot • Given n keys and m slots in the table, the load factor = n/m = average # keys per slot • What will be the average cost of an unsuccessful search for a key? A: O(1+)

Analysis of Chaining • Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot • Given n keys and m slots in the table, the load factor = n/m = average # keys per slot • What will be the average cost of an unsuccessful search for a key? A: O(1+) • What will be the average cost of a successful search?

Analysis of Chaining • Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot • Given n keys and m slots in the table, the load factor = n/m = average # keys per slot • What will be the average cost of an unsuccessful search for a key? A: O(1+) • What will be the average cost of a successful search? A: O(1 +/2) = O(1 + )

Analysis of Chaining Continued • So the cost of searching = O(1 + ) • If the number of keys n is proportional to the number of slots in the table, what is ? • A:  = O(1) • In other words, we can make the expected cost of searching constant if we make  constant

Choosing A Hash Function • Clearly choosing the hash function well is crucial • What will a worst-case hash function do? • What will be the time to search in this case? • What are desirable features of the hash function? • Should distribute keys uniformly into slots • Should not depend on patterns in the data

Hash Functions:The Division Method • h(k) = k mod m • In words: hash k into a table with m slots using the slot given by the remainder of k divided by m • What happens to elements with adjacent values of k? • What happens if m is a power of 2 (say 2P)? • Upshot: pick table size m = prime number not too close to a power of 2

What does this term represent? Hash Functions:The Multiplication Method • For a constant A, 0 < A < 1: • h(k) =  m (kA - kA) 

Hash Functions:The Multiplication Method • For a constant A, 0 < A < 1: • h(k) =  m (kA - kA)  • Choose m = 2P • Choose A not too close to 0 or 1; A=s/2w • Knuth: Good choice for A = (5 - 1)/2 Fractional part of kA

Hash Functions: Worst Case Scenario • Scenario: • You are given an assignment to implement hashing • You will self-grade in pairs, testing and grading your partner’s implementation • In a blatant violation of the honor code, your partner: • Analyzes your hash function • Picks a sequence of “worst-case” keys, causing your implementation to take O(n) time to search • What’s an honest CS student to do?

Hash Functions: Universal Hashing • As before, when attempting to foil an malicious adversary: randomize the algorithm • Universal hashing: pick a hash function randomly in a way that is independent of the keys that are actually going to be stored • Guarantees good performance on average, no matter what keys adversary chooses

CSC 413/513: Intro to Algorithms